Question

If the uncertainty in the position is equal to the wavelength of an electron , how certain can we be about the velocity. Determine an expression for $\dfrac{{{v_x}}}{{\Delta {v_x}}}$.

Answer 1

Heisenberg’s uncertainty principle is used to tackle this problem and in deriving the expression. De Broglie waves are applicable in the case of electrons and not photons because photons have no mass.

Complete step by step answer:
The Heisenberg uncertainty principle tells us that the product of uncertainty in position and uncertainty in momentum is greater than or equal to $\dfrac{h}{{2\pi }}$. That is,
$\Delta x\Delta p \geqslant \dfrac{h}{{2\pi }}$ −−−−− (1)
Where $h$ is the planck's constant.
The de Broglie wavelength is given by,
$\lambda = \dfrac{h}{p} \\\ \Rightarrow \lambda = \dfrac{h}{{m{v_x}}}$
When $\Delta x = \lambda$ , (1) becomes
$\dfrac{{{h}}}{{m{v_x}}}\times\Delta {p_x} \geqslant \dfrac{{{h}}}{{2\pi }}$ −−−−− (2)

Since $p = mv$ $\Rightarrow \Delta p = m\Delta v$
(2) becomes, $\dfrac{1}{{{m}{v_x}}}\times{m}\Delta {v_x} \geqslant \dfrac{1}{{2\pi }}$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow\dfrac{1}{{{v_x}}}\times\Delta {v_x} \geqslant \dfrac{1}{{2\pi }}$
$\Rightarrow$ $\dfrac{{\Delta {v_x}}}{{{v_x}}} \geqslant \dfrac{1}{{2\pi }}$
Taking the reciprocal on both sides,
$\dfrac{{{v_x}}}{{\Delta {v_x}}} \leqslant 2\pi$
The wavelength for an electron is of the order ${10^{ - 9}}$ , therefore uncertainty in the velocity is large for maintaining the inequality $\Delta x\Delta p \geqslant \dfrac{h}{{2\pi }}$.

Therefore, This is true because $\Delta {v_x}$ should be very large in order to make $\dfrac{{{v_x}}}{{\Delta {v_x}}} \leqslant 2\pi$.

Note: The uncertainty principle states that if we know the exact position of a particle then we have no information of its momentum and vice versa. The more we know about either of these values, the less we know about the other. In other words, A particle is dispersed over space, so it does not occupy a single definite location, but rather a range of positions. Similarly, because a particle is made up of a packet of waves, each of which has its own momentum, the momentum of a particle can only be described as a range of momentum at best.